What is the integral from 0 to 5 of sqrt(25+x^2) ?

The integral from 0 to 5 of sqrt(25+x^2) is (25(sqrt(2)+ln(1+sqrt(2))))/2

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integral from 0 to 5 of sqrt(25+x^2)

\int_{0}^{5} 25 + x^{2} d x

\frac{25 ( 2 + ln ( 1 + 2 ) )}{2}

Decimal Notation

28.69483 \dots

Solution steps

\int_{0}^{5} 25 + x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} 25 sec^{3} (u) d u

= \int_{0}^{\frac{π}{4}} 25 sec^{3} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 25 \cdot \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

Apply Integral Reduction

= 25 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u)

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= 25 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Simplify

25 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 25 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 25 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Compute the boundaries:

\frac{1}{2}

= 25 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

Simplify

= \frac{25 ( 2 + ln ( 1 + 2 ) )}{2}

\int \frac{sec ^{2} ( x )}{1 + tan ( x )} d x

n = 1 \sum \infty \frac{3 n}{4 n + 1}

\frac{( x + 2 ) ^{(2)} ( d y )}{( d x )} = 5 - 8 y - 4 x y

\int \frac{( y - 2 )}{y + 3} d y

\int_{0}^{0.6} 2 x e^{- x} d x

What is the integral from 0 to 5 of sqrt(25+x^2) ?
The integral from 0 to 5 of sqrt(25+x^2) is (25(sqrt(2)+ln(1+sqrt(2))))/2